Question 7 of 11.5 step 2 of 2
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A random sample of 9 fields of corn has a mean yield of 34.7 bushels per acre and standard deviation of 8.04 bushels per acre. Determine the $98\mathrm{\%}$ confidence interval for the true mean yield. Assume the population is approximately normal.

Step 2 of 2 : Construct the $98\mathrm{\%}$ confidence interval. Round your answer to one decimal place.
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Step 1 ：Given that the sample mean ($\overline{x}$) is 34.7 bushels per acre, the Z-score ($Z_{\alpha /2}$) for a 98% confidence interval is approximately 2.33, the standard deviation ($\sigma$) is 8.04 bushels per acre, and the sample size ($n$) is 9.

Step 2 ：We use the formula for the confidence interval which is given by: $$\overline{x}\pm Z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}$$

Step 3 ：Substitute the given values into the formula to calculate the confidence interval: $$34.7\pm 2.33\times \frac{8.04}{\sqrt{9}}$$

Step 4 ：The margin of error is calculated to be approximately 6.2444.

Step 5 ：The lower limit of the confidence interval is calculated as $34.7-6.2444=28.5$

Step 6 ：The upper limit of the confidence interval is calculated as $34.7+6.2444=40.9$

Step 7 ：The $98\mathrm{\%}$ confidence interval for the true mean yield is $\boxed{{\displaystyle (28.5,40.9)}}$ bushels per acre.